Steven Watanabe-4 wrote
AMDG
On 07/18/2012 06:57 AM, Larry Evans wrote:
On 07/18/12 01:29, Vicente J. Botet Escriba wrote:
Hi,,
In the example from the doc
http://steven_watanabe.users.sourceforge.net/type_erasure/libs/type_erasure/...
typedef mpl::vector< copy_constructible<_a>, copy_constructible<_b>, copy_constructible<_c>, constructible<_a(const _b&, const _c&)> > construct;
std::vector<double> vec; int i = 10; double d = 2.5; tuple<construct, _a&, _b, _c> t(vec, i, d); any<construct, _a> v(get<1>(t), get<2>(t));
I don't see which type is constructed and stored in v and how this type is given. That is I don't see how _a is binded. Could you help me?
I'd guess that a std::vector<double>&, more specifically the vec is stored as the _a& in the tuple<construct, _&a, _b, _c>. Hmm... Now I'm not sure anymore. Maybe it's a new std::vector<double> that's stored and the std::vector<double> CTOR args are 10 and double. IOW the equivalent of:
std::vector<double> v_a(10,2.5)
is stored in v?
Yes.
Could the documentation on this be clarified a bit?
What needs to be clarified? This is the only reasonable behavior: The arguments obviously match constructible<_a(const _b&, const _c&)> right? Therefore this constructor is the one called. _a was bound to std::vector<double>, so this is the type created.
In Christ, Steven Watanabe
My question is how the bind of _a to std::vector<double> if the t variable is not passed in the constructor of any? I'm sure I'm missing something evident. Vicente -- View this message in context: http://boost.2283326.n4.nabble.com/TypeErasure-Forward-constructors-and-bind... Sent from the Boost - Dev mailing list archive at Nabble.com.